A rough pipe is of diameter 8.0 cm. The velocity at a point 3.0 cm from the wall is 30% more than the velocity at a point 1 cm from the pipe wall. Determine the average height of the roughness.

 A rough pipe is of diameter 8.0 cm. The velocity at a point 3.0 cm from the wall is 30% more than the velocity at a point 1 cm from the pipe wall. Determine the average height of the roughness.

Solution:- 

Given:-

D=8m=0.08m

Let the velocity of flow at 1 cm from the pipe wall = u

Then the velocity of flow at 3 cm from pipe wall = 1.3u

uu∗${\displaystyle \frac{u}{u_{\ast }}}$ = Velocity distribution for rough pipe

uu∗=5.75log10(yK)+8.5${\displaystyle \frac{u}{u_{\ast }}}=5.75{\log }_{10}\left({\displaystyle \frac{y}{K}}\right)+8.5$

a) For a point, 1 cm from pipe wall

uu∗=5.75log10(1K)+8.5………..(1)${\displaystyle \frac{u}{u_{\ast }}}=5.75{\log }_{10}\left({\displaystyle \frac{1}{K}}\right)+8.5\dots \dots \dots ..(1)$

(b) For a point, 3 cm from pipe wall, velocity is 1.3 u, and hence

1.3uu∗=5.75log10(3K)+8.5…………(2)${\displaystyle \frac{1.3u}{u_{\ast }}}=5.75{\log }_{10}\left({\displaystyle \frac{3}{K}}\right)+8.5\dots \dots \dots \dots (2)$

Dividing (1) and (2) equation, we get

1.3=5.75log10(3K)+8.55.75log10(1K)+8.5$1.3={\displaystyle \frac{5.75{\log }_{10}\left({\displaystyle \frac{3}{K}}\right)+8.5}{5.75{\log }_{10}\left({\displaystyle \frac{1}{K}}\right)+8.5}}$

1.3[5.75log10(1K)+8.5]=5.75log10(3K)+8.5$1.3\left[5.75{\log }_{10}\left({\displaystyle \frac{1}{K}}\right)+8.5\right]=5.75{\log }_{10}\left({\displaystyle \frac{3}{K}}\right)+8.5$

7.475log10(1K)−5.75log10(3K)=8.5−11.05$7.475{\log }_{10}\left({\displaystyle \frac{1}{K}}\right)-5.75{\log }_{10}\left({\displaystyle \frac{3}{K}}\right)=8.5-11.05$

7.475log10(1K)−5.75log10(3K)=−2.55$7.475{\log }_{10}\left({\displaystyle \frac{1}{K}}\right)-5.75{\log }_{10}\left({\displaystyle \frac{3}{K}}\right)=-2.55$

7.475[log10(1.0)−log10(K)]−5.75[log10(3.0)−log10(K)]=−2.55$7.475[{\log }_{10}(1.0)-{\log }_{10}(K)]-5.75[{\log }_{10}(3.0)-{\log }_{10}(K)]=-2.55$

7.475[0−log10(K)]−5.75[0.4771−log10(K)]=−2.55$7.475[0-lo{g}_{10}(K)]-5.75[0.4771-{\log }_{10}(K)]=-2.55$

−7.475log10(K)−2.7433+5.75log10(K)]=−2.55$-7.475lo{g}_{10}(K)-2.7433+5.75{\log }_{10}(K)]=-2.55$

−1.725log10(K)=2.7433−2.55$-1.725lo{g}_{10}(K)=2.7433-2.55$

−1.725log10(K)=0.1933$-1.725lo{g}_{10}(K)=0.1933$

log10(K)=0.1933−1.725$lo{g}_{10}(K)={\displaystyle \frac{0.1933}{-1.725}}$

log10(K)=−0.1120=1¯¯¯.888$lo{g}_{10}(K)=-0.1120=\overline{1}.888$

K=0.7726 cm